62 lines
2.4 KiB
Markdown
62 lines
2.4 KiB
Markdown
## \-\-- Day 23: Coprocessor Conflagration \-\--
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You decide to head directly to the CPU and fix the printer from there.
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As you get close, you find an *experimental coprocessor* doing so much
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work that the local programs are afraid it will [halt and catch
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fire](https://en.wikipedia.org/wiki/Halt_and_Catch_Fire). This would
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cause serious issues for the rest of the computer, so you head in and
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see what you can do.
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The code it\'s running seems to be a variant of the kind you saw
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recently on that [tablet](18). The general functionality seems *very
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similar*, but some of the instructions are different:
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- `set X Y` *sets* register `X` to the value of `Y`.
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- `sub X Y` *decreases* register `X` by the value of `Y`.
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- `mul X Y` sets register `X` to the result of *multiplying* the value
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contained in register `X` by the value of `Y`.
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- `jnz X Y` *jumps* with an offset of the value of `Y`, but only if
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the value of `X` is *not zero*. (An offset of `2` skips the next
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instruction, an offset of `-1` jumps to the previous instruction,
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and so on.)
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The coprocessor is currently set to some kind of *debug mode*, which
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allows for testing, but prevents it from doing any meaningful work.
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If you run the program (your puzzle input), *how many times is the `mul`
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instruction invoked?*
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Your puzzle answer was `5929`.
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## \-\-- Part Two \-\-- {#part2}
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Now, it\'s time to fix the problem.
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The *debug mode switch* is wired directly to register `a`. You [flip the
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switch]{title="From 'magic' to 'more magic'."}, which makes *register
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`a` now start at `1`* when the program is executed.
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Immediately, the coprocessor begins to overheat. Whoever wrote this
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program obviously didn\'t choose a very efficient implementation.
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You\'ll need to *optimize the program* if it has any hope of completing
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before Santa needs that printer working.
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The coprocessor\'s ultimate goal is to determine the final value left in
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register `h` once the program completes. Technically, if it had that\...
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it wouldn\'t even need to run the program.
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After setting register `a` to `1`, if the program were to run to
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completion, *what value would be left in register `h`?*
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Your puzzle answer was `907`.
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Both parts of this puzzle are complete! They provide two gold stars:
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\*\*
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At this point, you should [return to your Advent calendar](/2017) and
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try another puzzle.
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If you still want to see it, you can [get your puzzle
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input](23/input).
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